It is known that the quadratic equation of one variable (m-radical 5) x square + 3x + m square-5 has a root of 0. Find the value of M

It is known that the quadratic equation of one variable (m-radical 5) x square + 3x + m square-5 has a root of 0. Find the value of M

(m-radical 5) x square + 3x + m square-5 has a root of 0
It shows that X1 = 0
M square - 5 = 0
M = positive and negative root sign 5
Because it's a quadratic equation of one variable
M = negative root 5

To solve the quadratic equation of one variable: (1) 12x2-x + 6 = 0 (2) (x-radical 3) 2 = 4x + 12-4 radical 3 (3) (3x + 2) 2 = 4 (x-3) 2 The "2" after the brackets is quadratic

（1）12x²-x+6=0 ²
△=1-24*12

The quadratic equation with (1 + 2 radical 3) / 2, (1-2 radical 3) / 2 as roots

x²-x-11/2=0
2x²-2x-11=0

Write a quadratic equation of one variable, so that its roots are 2 + radical 3 and 2-radical 3 As above

Xsquare - 4x + 1 = 0

The quadratic equation with roots 5-1 / 2 and 5 + 1 / 2 as roots is

The sum of the two is
√5-1/2+√5+1/2=2√5
The product of two is
(√5-1/2)*(√5+1/2)=5-1/4=19/4
therefore
The equation is
x²-2√5x+19/4=0

A quadratic equation with a root of 2 and a negative root sign 2 can be

（x-2）（x-√2）=0
x²-（2+√2）x+2√2=0
The quadratic equation of one variable with 2, negative root sign 2 as root can be x 2 - (2 + √ 2) x + 2 √ 2 = 0

Is there a quadratic equation of one variable with root 2, - radical 2? As the title

（x-2)(x+√2）=0

Write a quadratic equation of one variable so that its two roots are root 2 and root 2 respectively

0

Please write a quadratic equation of one variable about X whose root is x = 1 plus or minus root 2

X^2-2X-1=0

Solving the definition domain of sine and cosine function in senior one mathematics y=2-sinx y=lg(cosx) Y = SiNx under radical + 25-x ^ 2 under radical

y=2-sinx
The domain of SiNx is r
So the domain defined here is also r
y=lg(cosx)
cosx>0
So the domain is 2K π - π / 2